std::move() is just casting?

Question

I have read a few (pesudo) implementions of std::move(). And all are just casting away the reference of the parameter and then returning it as a rvalue reference.

It do

Answer 1

Yes, that's exactly as it is described in the standard. In N4659 (which is last draft I found)

it says in §23.2.5

template constexpr remove_reference_t&& move(T&& t) noexcept;

Returns: static_cast&&>(t)

It doesn't mark anything for destruction, and it doesn't change the object but object may be changed in function that accepts rvalue (such as move constructor, move assignment operator)